How do you differentiate #f(x)=(sinx+lnx)(x-3e^x)# using the product rule?
1 Answer
Jan 19, 2016
Explanation:
The product rule states that if
#f'(x)=g'(x)h(x)+g(x)h'(x)#
In this instance, we have
#g(x)=sinx+lnxcolor(white)(xxxx)=>color(white)(xxxx)g'(x)=cosx+1/x#
#h(x)=x-3e^xcolor(white)(xxxx)=>color(white)(xxxx)h'(x)=1-3e^x#
Thus, we have
#f'(x)=(cosx+1/x)(x-3e^x)+(sinx+lnx)(1-3e^x)#
